In the reaction of sodium thiosulphate with l₂ in aqueous medium, the equivalent weight of sodium thiosulphate is equal to

$2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3 + {\mathrm{I}}_{2 }⟶ 2\mathrm{NaI} + {\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$

Equivalent weight is the molecular weight divided by number of electrons lost or gained by each molecule.

Half reactions occur as:
$$\begin{gathered} {\mathrm{I}}_2 + 2{\mathrm{e}}^{-}\to 2{\mathrm{I}}^{-} \\ 2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}⟶{\mathrm{S}}_4{\mathrm{O}}_6^{2-}+2{\mathrm{e}}^{-} \end{gathered}$$
2 electrons are lost by 2 molecules of  S₂​O₃​²⁻(Sodium thiosulphate). So, one molecule will lose 1 electron. So, equivalent weight is equal to molecular weight divided by 1.

$\therefore$Equivalent weight of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$
$=\frac{\text{ Mol.wt. }}{1}=\frac{248}{1}=248$